in this video we're going to focus on synthesis reactions reactions associated with alkes uh gred reagents and we're going to cover some other examples but let's begin so this is acetylene which is basically a two carbon alkine and what we're going to do first is we're going to add sodium Ami to it and a nh2 the nh2 ion is a very strong base it's strong enough to remove the hydrogen from the alkine the pka of an alkine is relatively acidic it's about 25 compared to most carbon hydrogen bonds so once we remove the hydrogen we now have what is known as an acetal ion and whenever carbon has a negative charge it's relatively nucleophilic and so we can react with an alcohal because bromine is more electr negative than carbon it pulls the electrons toward itself and so bromine has a partial negative charge and therefore carbon has a partial positive charge when you have a positively charged carbon atom and another one that's negatively charged these two carbon atoms will connect and they will create a carbon carbon Bond so this carbon with negative charge is attracted to the carbon with a positive charge and so the two will connect and that's how we can make carbon carbon bonds and so this is the product for this reaction so I just want to give you an overview of the mechanism of that reaction but now let's say if you have this Aline what reagents do you need to make this product how would you propose a list of reagents that you need to make that and feel free to write the mechanism as well now notice that we have two carbons already but right now we have on the right side six so we need to add four carbons so basically we need to add this part of the molecule so what we need first is we need sodium amide to remove the hydrogen and then second we need an aloh halide basically a four carbon alcohal and then third we need to reduce the alkine into an Aline and what we could use to do that is um we can use a sodium metal with uh liquid ammonia so now let's go over the mechanism for this process so in the first step the amide ion is going to remove a hydrogen generating the acetal ion and then in the next step this is going to react with let me redraw that better this is going to react with one bromobutane so this is the carbon that has the partial positive charge it's the carbon that is directly attached to the bromy atom and so the acetal ion is going to attack the carbon and it's going to expel the bromine atom and it does so by means of an S2 reaction and so at this point we have a total of six carbon atoms now once we add sodium metal with liquid ammonia this is going to reduce the the triple bond or the alkine into an alken and so that's how we get this product so let's try another example let's say if uh you have this alkine and so St from acetylene you want to make this product how would you do it so notice that the two carbons of the triple bond are basically these two carbons with a double bond so we need to add an R Group for the left side and for the right side which we can do because there's two hydrogens and we can remove both of them so first let's add these two carbons step one we need to add sodium amide to remove the hydrogen step two we're going to add two carbons and a br so basically a two carbon alcohal or in this case ethol bromide now step three we need to use uh sodium amide again to remove the second hydrogen and then in step four we're going to add this part of the molecule so basically just redraw what you see we need a total of five extra carbons and then we're going to add a leing group to it now how can we convert a triple bond into a CIS double bond or CIS alen to convert the alkine into a cyst alkan you need to use hydrogen gas but with the lender's catalyst the lindler catalyst is basically um a deactivated padium catalyst so you might see it as a padium uh poison with like barium sulfate or something or with some other ionic compound so it's just a modified or deactivated plaum catalus and it's going to give you the CIS Aline another way in which we can make a carbon carbon bond is by using a gred reagent so here's an example of a gred reagent what we have here is methol magnesium bromide and we know oxygen has a partial negative charge it's more Electro negative than carbon but the carbon that is attached to the oxygen has a partial positive charge now the ch3 that's attached to the metal to magnesium that carbon has a partial negative charge so whenever you have a nucleophilic carbon and an electrophilic carbon those two will react to create a carbon carbon Bond and so initially we're going to get this Al oxide ion and we're going to add a ch3 to it now the magnesium and the bromide they're ionically bonded to the uh oxygen with the negative charge now once we acidify the solution using uh H2O plus we can now protonate the ALCO oxide ion and so we're going to get a tertiary Al so that's the uh Gren reagent and in your organic chemistry course you might see a few examples um dealing with this reaction so starting with cyclopentanol how can you make this product what reagents do you need now you can use a gred reagent or an Organo lithium reagent but let's use a Gren reagent so we need to add four carbons uh to this compound so basically we need four carbons and an mgbr group in the second step we need h2+ that's it that's all the reagents you need to make that product now whenever you add a grin reg agent to an alide or Ketone only one R Group will be added but let's say if you have like an acid chloride or an Esther you can add two R groups so for example let's say if you were to use a methyl magnesium bromide followed by h2+ this will be reduced to an alcohol but by the addition of two methyl groups so let's see why that's the case so starting with this acid chloride the grenade reagent is going to attack the Carbono carbon and so we're going to add a ch3 to it and then when that happens this this oxygen is going to regenerate the double bond or the pi Bond and it's going to expel the chlorine uh the chlorine atom the chlorine is a good leaving group so it's very easy to kick it out and so right now we have a ketone how now at this point the grenard can still react with the keto so another Griner re agent that's in a solution will continue in this reaction it's going to reduce the Ketone as well so now we have an oxygen with a negative charge and at this point we have two methyl groups so the reaction stops until we acidify the solution with uh hydronium ions so in the end we're going to get a tertiary alcohol so the Griner reg agent is very useful for uh making alcohols and it does so by adding carbon atoms so this is the answer for this uh reaction or example so starting from an acid chloride or even an Esther how can we convert the acid chloride into a ketone we know that a Griner reagent won't stop at the Ketone level it's going to take it all the way to an alcohol but what if we wanted to stop at the Kon level or what if we want to take an Esther and convert it into the uh alahh level how can we do that well if you want to make a ketone what you need to use is the the Gilman Regent or the Organo lithium Organo copper lithium Regent this will reduce the acid chloride to a ketone but it won't react with the Ketone and if you want to get the alahh you need to use something called Dao so I'm going to go over the mechanism for these processes so starting with the acid chloride here's the uh the Gilman reagent there's two methyl groups attached to the copper atom and there's a lithium ion that's next to it so one of the methyl groups attacks the acid chloride but not both just one and at this point one of the lone paers will be used to regenerate the pi Bond expel in the chloride Le group and the reaction stops here only one of the meth groups is nucleophilic once you get rid of the methyl group The Copper will lose this negative formal charge and so it's going to be strongly attached to this the the remaining methyl group so only one of the meth groups is nucleophilic but what you need to keep in mind is for the go reg agent if you combine B it with an acid chloride or anther it's going to stop at the Ketone level and that's what you want to take from this reaction so now let's go over the mechanism between an Esther and uh daba so daba it's basically an aluminum atom that has only one hydrogen atom but also has like three o groups now the aluminum atom has a negative formal charge and lithium has a positive charge so whenever hydrogen is attached to a metal it's going to have a partial negative charge it's nucleophilic kind of like when a carbon is attached to a metal keep in mind formal charge and partial charge are not the same thing they're different now this hydride is going to attack the carbon and then the pi bond is going to break and just like before the carbonal group is going to be regenerated by the exposion of this leaving group now the oc3 is not a great leaving group but it can still leave and so the end result of this reaction is an alide it stops at this level so let's say if you have an acid chloride and you want to make make this particular product how would you go about doing it so feel free to pause the video and see if you can find the right reagents to make this product now what you want to focus on you want to think about what you need to add to make this product so before notice that you have a chlorine and it left but notice the groups that are attached to this carbon atom let's focus on that carbon so these three carbons were already there we didn't add it we added a benzing ring but notice that there is an invisible hydrogen that is now visible since I wrote it there so basically we added a hydrogen and a Benz ring now we can't add a grin reagent right now to this acid chloride because it's going to add two R groups basically two R groups that contain carbon so we want to add a hydrogen so we don't want use the grin right now so the first thing you want to do to add the hydrogen is you want to use Dao it's going to take it down to the alide level by adding a hydrogen and now you don't want to use the Gil Regent at this point it doesn't really react with ketones and I haven't seen it react with alahh but what you want to use now is the gred reagent or you can use um an Organo lithium reagent so you can use ch3 mgbr at this point well not ch3 but we need to add the benzing ring so you can use a phenol magnesium bromide followed by h2+ that's the Gren agent or you can use uh this reagent and or again a lithium reagent with a Benz ring so both of these can work so and that's going to be the answer so step one really is using daone to add the hydrogen once you do that step two is to add Phenom magnesium bromide and then step three H2O plus and that's how you can make it so now let's say if we have uh an Esther and whatever agents do we need to make uh this particular product how would you make it so we need to add two different R groups and how would you make this one as well so I want to highlight the difference here notice that for the uh product on the bottom we're adding the same R groups so if you want to add two of the same type of R groups simply use a Gren reagent so the reagent that we need is basically Ally this grer agent followed by uh H2O plus now in the first example on top we want to add two different R groups so you don't want to use a grin reagent initially you want to do something else what you want to use is the Organo lithium copper reagent so let's use uh ch3 ch2 times to coer lithium what's going to happen is it's going to take it down to the Ketone level it's going to replace the oc3 with one of these R groups so initially you're going to get uh this product so now that it's at the Ketone level you can now add the grity reagent to add the second R Group so we're going to use uh fenel uh magnesium bromide and that's going to add the second R Group which is different from the first first and then step three we're going to add h2+ so that's what you can do if you wish to add two different R groups to an estor or an acid chloride so first use a g v agent for the first R Group and then use a grer v agent for the second R Group and then you can get it done now if you want to add uh two R groups that are identical to each other you can simply use the gr reg agent and you can just write two equivalents of that reagent to uh to get this product if you want to get it in good yield now let's talk about how some of these reagents are made starting with an aloh halide all you need to do to make a Gren reagent is simply uh add a metal to it particularly magnesium metal magnesium is going to insert itself between uh the carbon and a bromine atom so basically you could see it this way this much help magnesium has two valence electrons and it's going to give the carbon atom those valence electrons right now carbon is partially positive it's directly attached to an electronegative atom and so once magnesium gives those two electrons to carbon carbon has no need of being attached to bromine so right now we have a carbon that has a negative charge and since magnesium lost two electrons it has a plus two charge and bromide it pull those two electrons in this Bond so it has uh an extra lone pair so it has negative charge so now we can combine imagine if this carbon with the negative charge attack the mg what we would have is something that looks like this the Magnesium will now have a plus one charge if you add negative 1 and positive2 you get plus one now imagine if the bromide ion attacks the Magnesium so a plus and a minus they pretty much will neutralize each other and so you get this compound so that's just one way to see how this works it may not be the exact mechanism but it helps you to understand how uh these atoms are like bonded to each other now what about making the Organo lithium reagent now what you need is two equivalents of lithium because lithium has only one valence electron whereas magnesium has two and to get rid of that bromine you need two electrons so I'm going to draw two lithium atoms each with one balanced electron so this lithium is going to contribute one electron and this one is going to contribute the other electron and so bromine is going to leave whenever you have a half Arrow it represents the flow of one electron a full Arrow represents a flow of two electrons and so now we have a carbon atom with um a negative charge we have two lithium ions and we also have a bromide ion so this carbon with a negative charge canare it with the lithium and so when you add a negative and a positive the net charge is zero so this is going to be the Organo lithium reagent so notice that there only one lithium atom attached to it but no bromine atom now these two can get together and form an ionic bond and so these are the products of this reaction so this reagent I'm going to represent it as a r Li now if you take this Organo lithium agent pretty much if you have like two of them you can react it with uh copper chloride copper 1 chloride specifically which I'm going to write it like this so one of the r groups will attack the copper kicking out the CL so now you have R cuu and the lithium has a plus one charge and chloride has a minus one charge so these two will get together and form l i now the second R lithium compound will also attack uh the copper now keep in mind though the R Group has a negative charge lithium has a plus charge so if we add this extra R Group we're adding basically a negative charge to this compound right now the compound has a net charge of zero but once we add this negative one uh charge to it it's going to have an overall charge of minus one so now that carp has two R groups it has a formal charge of negative 1 and so it's going to pair up with the lithium and so thus we can write it as R2 copper lithium so when you uh take an Organo lithium agent you need two equivalents and when you react it with like copper chloride you're going to get uh R2 copper lithium plus l i so that's how you can make the Gilman Regent so let's say if you have an aloh helide the first thing you could do is add lithium to it and this is going to turn into uh this compound and then once you add copper chloride assuming you add enough of this let's say two equivalents then you can get the the R2 copper lithium reagent now what can we do with this we can react this group with another alalal notice what's going to happen so if we start with uh one bromobutane and if we reacted with an Organo copper lithium uh reagent this R Group will attack this carbon and expel the BR so basically we can couple uh two different alkal groups together so here we had four carbons and here we also have four carbons so we're going to get an eight carbon structure so that's how you can combine uh two alkal groups together so let's say if we have a a f like a a five um carbon alcohal and uh a four carbon alcohal so step one if we add lithium to one of these alcohal lies and then step two if we add uh copper chloride to it it and step three if we add this thing we can connect these two together and so we're going to get nine carbons and so basically we're just combining these two chains together so that's how you can combine two alcohal lies so let's say if we have the following alcohal what reagents do you need to make this product how would you do it so let's count the carbons that we already have 1 2 3 four five I'm counting the longest chain we also have two additional methyl groups in Carbon 2 so if I count the same carbon atoms 1 2 3 4 five I know that this structure is already present in my reactant so what I need to add to it is the part that remains which is this part so step one I'm going to convert this alcohal into an Organo lithium reagent so step one I'm going to add lithium and then I'm going to convert it into a g reagent so I'm going to add a copper chloride and then in step three once I have my G rent I'm going to add the other alcohal that I need which is basically this one so I need an alcohal that has bromine on this carbon so I'm going to draw it backwards so I'm going to start with this uh the two method groups at the end basically like an isopropyl group so let me count the carbons 1 2 3 four five so on the fifth carbon I need a bromine atom so one two three four five and then BR so those are the reents I need to make uh this particular product but let's go through through each of the steps so step one was the addition of lithium which will make the or uh the Organo lithium agent and then step two was adding copper chloride now granted I need to use two equivalents so let's say if I have um eight moles of this product all I need is four moles of copper chloride so I need like a 2:1 ratio so to speak and so I'm going to have two of these R groups and then uh copper lithium attached to it so now I can react it with the other aloh halite that I had and so what's going to happen is one of the r groups is going to attack this carbon and expel the BR and then so the product that I'm going to get looks like this and so this will be my answer I believe I counted the carbons correctly but let me make sure yeah this should be right and so that's how you can combine two aloh light together if you ever need to do so now there's some other cool things that you can do with Organo lithium reagents so let's say if you have an alen and attached to this alkane is like a halogen on it you can add an organocopper lithium agent to it and and what's going to happen the net result is that you're going to replace the bromine atom with a methyl group and a double bond is still going to be there so here's how it works so one of the methyl groups is going to attack the carbon that has the bromine atom because keep in mind that carbon has a a significant amount of partial positive charge but it's not like an sn2 reaction where it attacks the car carbon and expels the bromine atom at the same time so this doesn't happen because it doesn't approach the molecule from the back so or at least not directly from the back so it's not easy that that doesn't happen so what happens instead is as the methyl group attacks the carbon the pi Bond breaks so it's not really like an sn2 reaction but the overall effect is a substitution process so the bromine atom is still here here initially at this point and the methyl group has been added so the first step was an addition step and then the second step is going to be an elimination step the carbon with the negative charge is going to use the lone here to regenerate the double bond and at it is at that point that it's going to expel the bromine atom and so the second step is an elimination step so this is called like an addition elimination reaction which combined overall is still considered a substitution reaction because in the end we replace or substituted a bromine atom with a methyl group so once that lone pair forms double bond and a bromine atom is expelled now we're going to have the product which looks like this so the Gman reagent is very useful for replacing halogens with r groups now you can also do something similar with uh a Benzene ring so if you were to add a a gilid reagent to this you can replace the bromine atom with a methyl group so this can work as well and the mechanism is very similar uh to the last example so starting from an alkane how can you make an alen and also how can you make an alkine and from an alkane how can you make a secondary alcohol and also a primary alcohol and how can you make an alahh and and a ketone and how can you get let's say an amine as well as a carboxilic acid or even an Esther or what about an ether so we're going to talk about how to get these different functional groups starting from an alkane so let's begin right now with an Alcan you can't pretty you can't really do much so the only thing you can do is probably add a bromine adom to it and that's usually the first thing you should do and one of the best ways to do it is to add NBS now you may have to add like heat or light or some sort of radical initiator but once you add NBS it's going to replace the most substituted hydrogen with a bromine atom so in this case bromine is selective it's going to repl the secondary hydrogen uh faster than it would replace the primary hydrogen so if you had a primary secondary or tertiary hydrogen the bromine atom will selectively replace the tertiary hydrogen so more than 99% of the product will be in this form you might get a small amount of one bromobutane but for the most part more than 99% is basically going to be like two bromobutane so you get this product in a very high yield now once we have two bromobutane we can add sodium hydroxide which is going to be an E2 elimination reaction and in this reaction the hydroxide ion is going to uh grab a hydrogen form a double bond and expel the lium group so that's how we can make the alken and once you have the alken you can do a lot of things so with the alen you can easily make a secondary alcohol or even a primary alcohol now to make this secondary alcohol Dev varients that you can use is uh Mercury acetate with H2O uh followed by sodium borohydride this is the oxy mercuration Dem mercuration reaction and it proceeds with marar the carve Edition now if you want to make the uh primary alcohol you can use uh bh3 borine with thf followed by hydrogen peroxide and hydroxide so this will give you the anti conar product so this is the hydroboration oxidation reaction now if you were to add a PCC the alcohol will be oxidized into a ketone and if you add PCC to the primary alcohol it's going to be oxidized into an alahh now if you were to use a strong oxidizing agent to this primary alcohol like chromic acid or potassium magnate with H2O plus you can convert it into a carboxilic acid now once you have the carboxilic acid you can do a lot of things so starting from the carboxilic acid you can add s SO2 and convert it into an acid chloride and once you have the acid chloride you can react it with let's say ammonia and that's going to turn into an amide with uh HCL being a byproduct you can also add an alcohol if you add an alcohol htl will be removed and you're going to get an Esther so basically you're going to replace the chlorine with this part of the molecule and so an acid chloride plus an alcohol will give you an Esther and and let's say if you were to add a carbic acid to it the chloride will be replaced with this part of the molecule basically you just have to get rid of the hydrogen and the chlorine so htl is going to be the byproduct so this will give you an an hydde now starting back with the Ketone and the alahh uh that we had in the uh previous examples once you have the ketone um you can add like ammonia to it and you're going to get an amine which looks like this if you add ammonia to this uh alahh you'll also get an IM but it's going to be on a primary carbon instead now you can reduce the IM with uh cyan borohydride so sodium cyano borohydride or na bh3 CN if you use it on a secondary Aman you're going to the double bond is going to be reduced to a single Bond so you're going to get a primary amine so that's going to be the product for this reaction and for this one the aiming is going to be on the primary carbon as opposed to the secondary carbon so here it's on the secondary carbon but this is still considered a primary aming because the nitrogen only has or is only attached to one carbon so this is a primary Aman but it's on a primary carbon so as you can see you can get both types of Ames um by reason of that reaction now going back to the alken we can make an ether if we use the AL coxy mercuration Dem mercuration reaction so if we were to use water we're going to get an alcohol but if we use an alcohol instead like um ch3oh instead of water instead of getting an O group we're going to get an oc3 group on the more subsidio carbon that's how you can convert an alen into an ether and you know how to convert the alkane into an alen now to get like a primary ether or an ether on the primary carbon once you have this alen you can add hbr with uh peroxide H2O2 and so this is going to put the bromine atom on the less substitute of carbon this reaction proceeds with anti M addition and now you can add sodium oxide when you use methoxide with a primary aqual halide the predominant mechanism is the2 reaction and so the methoxide is going to attack the carbon from the back expelling the bromine atom and so you're going to get an ether on a primary carbon as opposed to uh the secondary carbon so now starting with the uh alken how can we convert it into an alkine how would you do that so the first thing you need to do is you need to add br2 so this reaction is going to put two bromine atoms across the double bond and then at this point you can use a strong base but we're going to use a sodium amide so the nh2 minus ion is a very powerful base and first thing is going to do is it's going to grab one of the hydrogens and then a double bond is going to be formed and one of the BR is going to leave so this is going to be an E2 reaction so right now we have a double bond with a one bromine atom still on it so now another nh2 minus uh ion it's going to remove this hydrogen it's going to form a triple bond and expel this br so that's how we're going to get the alkine that's how you can make it now starting from the alkine how can we make a ketone and an alahh from it there's some reagents that you should know regardless so to make the Ketone what you really need is uh mercury sulfate with water and suric acid now initially before you get the Ketone you're going to get something called an enol and the enol is going to toize into the Ketone now for the next one if you want to make the alahh you need to use a reaction that's similar to hydroboration but notice that there's only one hydrogen on a boron it's not three otherwise you'll get some interesting side products but other than that it's very similar to the hydroboration oxidation reaction now initially you're going to get an enol that looks like this and it's going to toize into the alahh form so the enol is the intermediate of these reactions but this reagent will generate the Ketone for the most part and this will produce the alahh if you have a terminal alkine that's an alkine at the end of the chain now starting with a ketone how can you convert a ketone into an alken how can you make this product if you want to convert a ketone into an alen the best way to do it is to use the witting reaction so you need to add this part to the illage so it's going to look like this and so that's what you need so with that in mind try this example starting from this Ketone find out the illid that you need to make this product so all you need to do is draw the ID and and then add this portion of it to it and just draw exactly what you see so that's how you can convert a ketone into an alken and you can make all sorts of alen using divic reaction so here's a question for you how can you turn uh butane and ethane how can you combine them to make cyclohexane so if you have butane and and ethane and you have some other reagents what reagents can you use to make it form a ring so feel free to pause the video think about it see if you can solve it and then when you're ready unpause the video to see if you can uh if you have the right answer now there might be more than one way of getting this done but here's uh one way to do it so let's start with um first let's clear the page let's start with butane what we're going to do first is ADD NBS and so we're going to replace a secondary hydrogen with a a bromine atom after that we're going to add sodium hydroxide so this is going to be an E2 elimination reaction and the hydroxide is going to uh get rid of one of the hydrogens actually I don't want to use hydroxide I take it back because it's going to give me the Z product I want the hofen product so I'm going to use a bulky base turit oxide so turpy oxide is going to go for the more accessible hydrogen that is the primary hydrogen and not the secondary one so we're going to get a terminal um alkan which is what I want so now that we have this terminal Aline what I'm going to do is use NBS again and it's going to replace the alyc hydrogen now at this point I can use any strong base turp oxide or hydroxide because hydroxide will easily remove this hydrogen so we can get a conjugated alkine so it's going to grab this hydrogen form a double bond and expel the bromine atom so now what I have is one to Reb a Dy now I'm not sure if you see where I'm going with this but in time you will now starting with ethane I'm going to use uh NBS again to get ethyl bromide and then I'm going to use sodium hydroxide which is going to be an an E2 reaction and that's going to give me an alkan particularly uh Ethan so now what I'm going to do is I'm going to combine these two together so I have one three badine and ethine what do you think is going to happen when you put these two together one three badine is a dying that is going to participate in the deal's Auto reaction the deal's Auto reaction is a 4 plus2 CYO addition reaction and so it's going to form a six membered ring which is going to look like this so one of the double bonds specially this one connects carbons one and six together so the pi electrons in that Bond is used to make uh this Sigma Bond and the double bond between three and four those P electrons are used to connect carbon four and five so there's only one double bond that's left over now since we wanted cyclohexane not cyclohexene what we could do is add H2 with a padium catalyst and it's going to convert the alen into an alcade so that's how you can make uh cyclohexane from butane and from ethane so now from cyclohexane how can you make Benzene this is going to be the last example for this video so what do you think we need to do well the first thing we need to do is add NBS so we're going to add a bromine atom or replace a hydrogen with a bromine atom in this ring and then once we do that we're going to add sodium hydroxide so the hydroxide is going to get rid of the bromine atom and it's going to form a double bond and we're going to repeat this process multiple times but right now what we have is cyclohexene so starting from cyclohexene we're going to add NBS again and NBS this time is going to replace the alyc hydrogen the hydrogen that's one carbon away from the double bond and so we're going to get this and then we're going to add hydroxide again so at this point hydroxide is going to get rid of this hydrogen form a double bond and expel the bromine atom so now we have two double bonds and so we're going to add NBS again and one of the alyc hydrogens will be replaced it really doesn't matter which one because at this point the two remaining hydrogens are both a LIC so we're just going to replace one of them with a bromine atom and then hydroxide is going to remove the other hydrogen for a double bond kicking out the BR so the driving force for this reaction is stability every time you add a new double bond it becomes more stable compared to the previous uh double bond so right now we have three double bonds so it's very stable so Benzene the double Bonds in the Benzene ring are more stable than the double Bonds in 13 cyoh hexad and the these two double bonds they're still conjugated so they're more stable than cyclohexene so this reaction can easily work because the driving force is stability we're going from a less stable compound to a more stable compound and when that happens it's easy to uh form that kind of product so that is it for this video so you have some examples on how to synthesize different compounds of course there are many other reactions out there um if you want more reactions do a YouTube search and check out my video on on organic chemistry 1 and two um reactions review uh study guide it has like a list of reactions and reagents that will help you for organic one and two so check that out when you get a chance and thanks for watching this video and have a great day